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Determine the pH of a .50 M solution of H2S04 and also determine the concentration of the sulfate ion (SO42-)

  • Chemistry -

    I don't know where you are in chemistry but this problem is not as easy as it looks. The problem is that H2SO4 is a strong acid (100% ionized) for the first H^+ BUT the second one is not completely ionized (k2 = 1.2E-2). As a result, the first H that comes off reduces the ionization of the second one even further so the pH is essentially determined by the first ionization. ]
    H2SO4 ==> H^+ + HSO4^-
    HSO4^- ==> H^+ + SO4^2-
    k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
    So you set up an ICE chart for k1 and k2.
    ..............H2SO4 ==> H^+ + HSO4^-
    equilibrium.....0.......0.50....0.50*see below.

    ..............HSO4^- ==> H^+ + SO4^2-

    k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
    Now substitute.
    (H^+) 0.50 from first equn + x from second.
    (SO4^2-) = x from second equn
    (HSO4^-) = 0.50-x frm second equn.
    0.012 = (0.50+x)(x)/(0.5-x) and solve for x. You will need to solve the quadratic OR you can do it by successive approximations.
    The (H^+) then is 0.5+x and x is the sulfate.

  • Chemistry -

    In the above I used k1 after saying H2SO4 had no k1. It doesn't. I should have said the first ionization which I referred to k1.

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